a pitching machine throws a baseball at a batter with an initial velocity of 22.0 m/s at an angle of 30 degrees to the horizontal. the ball is 1.60 m above the ground when it is pitched.
a) if the batter hots the ball at the same height as it was launched how long is the ball in the air on its way to the batter?
b)how far is the ball from the pitching machine when it is hit?
c) if the batter misses the ball, where dose the ball land, relative to the pitching machine? relative to the batter?
Vo = 22m/s @ 30o.
Xo = Hor. = 22*cos30 = 19.1 m/s.
Yo = Ver. = 22*sin30 = 11 m/s.
b. Range = Vo^2*sin(2A)/g.
Range = (22)^2*sin(60)/9.8 = 42.8 m. =
hor. distance.
a. Range = Xo * T = 42.8 m.
19.1 * T = 42.8
T = 2.25 s. = Time in air.
c. Tr = T/2 = 2.25/2=1.125 s.=Rise time.
hmax = ho + (Y^2-Yo^2)/2g.
hmax = 1.6 + (0-121)/-19.6 = 7.77 m.
Above gnd.
hmax = Vo*t + 0.5g*t^2 = 7.77 m.
0 + 4.9t^2 = 7.77
t^2 = 1.586
Tf = 1.26 s. = Fall time.
Dx=Xo*(Tr+Tf)=19.1*(1.125+1.26)=45.6 m.
do you go to Anderson CVI? :o
No, i've never heard of it.
I go to anderson :|
To answer these questions, we need to break down the ball's motion into its horizontal and vertical components using basic principles of projectile motion.
Step 1: Calculate the initial vertical velocity (Vy) and initial horizontal velocity (Vx) of the ball.
Given:
- Initial velocity (V) = 22.0 m/s
- Launch angle (θ) = 30 degrees
Using trigonometry, we can calculate the vertical and horizontal components of the initial velocity:
Vy = V * sin(θ)
Vx = V * cos(θ)
Vy = 22.0 m/s * sin(30°) = 11.0 m/s
Vx = 22.0 m/s * cos(30°) = 19.0 m/s
a) To find the time of flight (t) of the ball, we can use the equation:
y = Vy * t + (1/2) * g * t^2
Where:
- y represents the vertical displacement (change in height) = -1.60 m (negative because the ball is moving downward)
- g is the acceleration due to gravity = 9.8 m/s^2 (taking the positive value)
Rearranging the equation gives us a quadratic equation: (1/2) * g * t^2 + Vy * t + y = 0
Using the quadratic formula, we can solve for t:
t = (-Vy ± sqrt(Vy^2 - 4 * (1/2) * g * y)) / (2 * (1/2) * g)
Let's plug in the values to find t:
t = (-11.0 ± sqrt(11.0^2 - 4 * (1/2) * 9.8 * -1.60)) / (2 * (1/2) * 9.8)
t ≈ 1.45 s (ignoring the negative value)
Therefore, the ball is in the air for approximately 1.45 seconds on its way to the batter.
b) To find the horizontal distance traveled by the ball, we can use the equation:
x = Vx * t
x = 19.0 m/s * 1.45 s
x ≈ 27.55 m
Therefore, the ball is approximately 27.55 meters away from the pitching machine when it is hit.
c) If the batter misses the ball, the ball will continue its projectile motion until it hits the ground.
To find the vertical displacement (range) of the ball, we can use the equation:
y = Vy * t + (1/2) * g * t^2
Since we know the time of flight (t) is approximately 1.45 seconds, we can find the vertical displacement:
y = 11.0 m/s * 1.45 s + (1/2) * 9.8 m/s^2 * (1.45 s)^2
y ≈ 7.93 m
Therefore, the ball will land approximately 7.93 meters below the pitching machine. Relative to the batter, it will depend on where the batter was positioned.